How prove exists $z_{0}$ such $|z_{0}|=1$,and $|f(z_{0})|\ge\frac{1}{2^{n-1}}\prod_{j=1}^{n}(1+|a_{j}|)$ 
let $a_{1},a_{2},\cdots,a_{n}\neq 0$ be given complex numbers and
  $$f(z)=\prod_{j=1}^{n}(z-a_{j})$$
I need to show that there exists a complex number $z_{0}$ such that $|z_{0}|=1$ and 
  $$|f(z_{0})|\ge\dfrac{1}{2^{n-1}}\prod_{j=1}^{n}(1+|a_{j}|)\tag{1}$$

PS: I have solve this old problem(old IMO Shortlist problem)
$$f(z)=z^n+a_{n-1}z^{n-1} :+\cdots+a_{1}z+a_{0},a_{i}\in C$$,show that: there exsit $z_{0}\in C$,such $|z_{0}|=1,|f(z_{0}|\ge 1$
Proof:use the Lagrange indentity to get
$$f(z)=\sum_{i=1}^{n+1}\dfrac{\displaystyle\prod_{j\neq i}(z-z_{j})}{\displaystyle\prod_{j\neq i}^{n}(z_{i}-z_{j})}f(z_{j})\tag{2}$$
where $z_{1},z_{2},\cdots,z_{n+1}$ are such that $(z_{i})^n=1,i \in [1,2 \cdots,n+1]$. Then we have
$$\sum_{i=1}^{n+1}\dfrac{f(z_{i})}{\displaystyle\prod_{j\neq i}(z_{i}-z_{j})}=1$$
and note $$\prod_{j\neq i}(z_{i}-z_{j})=\prod_{j=1}^{n}(z_{i}-z_{i+j})=(n+1)z^n_{i}$$,so $$|\prod_{j\neq i}(z_{i}-z_{j})|=n+1$$
from $(2)$ we have
$$\sum_{j=1}^{n+1}|f(z_{i})|\ge n+1$$
so the old problem is by done!
But I use this methods  stuck the problem $(1)$.
This is 2013 China Team Selection problem exercise
 A: For the function $z^n-1$ we have equality in the statement.
Two observations:
(1) We can assume that $|a_j|\le1$ for every $j$.
Proof:
If $|z|=1$ then
$|z-a|=|z|\cdot|\bar{z}-\bar{a}|=|1-\bar{a}z|=|a|\cdot\left|z-\frac1{\bar a}\right|$, and
$1+|a|=|a|\cdot\left(1+\frac1{|\bar{a}|}\right)$. If we replace the factor $z-a$ in the polynomial by $\left(z-\frac1{\bar a}\right)$, we obtain an equivalent statement.
(2) We can assume that the constant term in $f$ is a nonnegative real number, so it is $|a_1a_2\ldots a_n|$.
Proof: $f(z)$ can be replaced by $w^nf(\bar{w}z)$ with a suitable unit $w$.
Now let
$f(z)=c_nz^n+c_{n-1}z^{n-1}+\ldots+c_1z+c_0$ where $c_n=1$ and $c_0=|a_1a_2\ldots a_n|$,
and let $w_0,w_2,\ldots,w_{n-1}$ be the $n$th roots of unity. Then
$$
\frac1n \sum_{k=0}^{n-1} f(w^k) =
\frac1n \sum_{k=0}^{n-1} \sum_{j=0}^n c_j (w^k)^j = 
\sum_{j=0}^n \frac{c_j}n \sum_{k=0}^{n-1} (w^j)^k = 
c_0+c_n = 1+|a_1\cdots a_n|.
$$
Therefore,
$$
\max_{|z|=1}|f(z)| \ge \left|\frac1n \sum_{k=0}^{n-1} f(w^k)\right| =
1+|a_1\cdots a_n|.
$$
For $x,y\le 1$ we have $x+y<1+xy$; this leads to
$$
\prod_{j=1}^n (1+|a_j|) \le 2^{n-1} \big( 1+|a_1\cdots a_n| \big).
$$

The IMOSL problem can be solved by
$$
\frac1{2\pi}\int_0^{2\pi} f(e^{it}) e^{-nit} dt = 1
$$
(coefficient formula)
or
$$
\frac1{2\pi}\int_0^{2\pi} \big|f(e^{it})\big|^2 dt = 1 + |a_0|^2+\ldots+|a_{n-1}|^2
$$
(Parseval's formula).
